8.9 Powers of a Matrix
The Cayley-Hamilton Theorem:
An n × n matrix A satisfies its characteristic equation.
example:
A=
6 −2
6 −1
Characteristic Eq. :
λ 2 − ( trace ( A) ) λ + det ( A ) = 0
λ 2 − 5λ + 6 = 0
by Cayley-Hamilton : A2 = 6 −2 6 −2 = 24 −10
6 −1 6 −1
30 −11
A2 − 5 A + 6 I = 0
−5 A = −5
6 −2
6 −1
6I = 6
1 0
0 1
=
=
A2 − 5 A + 6 I =
−30 10
−30
5
6 0
0 6
0 0
0 0
Find a specified power of a matrix A :
Method 1: Using C-H Theorem and repeated multiplication
Section 8.9
example:
A=
6 −2
6 −1
Find A6 .
λ 2 − 5λ + 6 = 0
A2 − 5 A + 6 I = 0
Solve for A2 .
Multiply by A.
Plug in for A2 .
Multiply by A.
A2 = 5 A − 6 I
Plug in for A2 .
Multiply by A.
2
Plug in for A .
A3 = 5 A2 − 6 A
A3 = 5 ( 5 A − 6 I ) − 6 A A3 = 19 A − 30 I
A4 = 19 A2 − 30 A
A4 = 19 ( 5 A − 6 I ) − 30 A A4 = 65 A − 114 I
A5 = 65 A2 − 114 A
A5 = 65 ( 5 A − 6 I ) − 114 A
A6 = 211A2 − 390 A
Plug in for A2 . A6 = 211( 5 A − 6 I ) − 390 A
A5 = 211A − 390 I
Multiply by A.
A6 = 665
6 −2
6 −1
− 1266
1 0
0 1
=
A6 = 665 A − 1266 I
2724 −1330
3990 −1931
1
Section 8.9
Find a specified power of a matrix A :
Method 2: Using C-H Theorem and a system of equations
example:
A=
6 −2
6 −1
λ 2 − 5λ + 6 = 0
A2 − 5 A + 6 I = 0
Since A2 = 5 A − 6 I , every multiple of A will be
Find A6 .
Am = k1 A + k0 I , where k0 and k1 are real numbers.
As we saw in the previous slide, m = 6 k1 = 665 and k0 = −1266
Goal: To set up a system of equations and solve for k1 and k0 .
Am = k1 A + k0 I is a result of the char. eq. so C-H in reverse says each λ satisfies
λ m = k1λ + k0 . For this A, λ1 = 3 and λ2 = 2, and we want m = 6.
=
=
36
26
+
+
3k1
2k1
−2 R1 + R2
k0
k0
3
1
|
729 − R2 + R1
1
0
|
665
2
1
|
64
2
1
|
64
1
0
|
665
0
1
|
−1266
k1 = 665 and k0 = −1266
Find the inverse of a matrix using the Cayley-Hamilton Theorem
example:
λ 3 − ( trace ( A ) ) λ 2 + ( C11 + C22 + C33 ) λ − det ( A) = 0
2 0 1
−2 3 4
−5 5 6
sum of the diagonal cofactors
λ − 11λ + ( −2 + 17 + 6 ) λ − 1 = 0
3
2
C-H
−1
λ 3 − 11λ 2 + 21λ − 1 = 0
A3 − 11A2 + 21A − I = 0
Solve for I .
I = A3 − 11A2 + 21A
Multiply by A−1.
A−1 = A2 − 11A + 21I
5
8
2
0 1
1 0 0
−1
A = −30 29 34 − 11 −2 3 4 + 21 0 1 0
−50 45 51
−5 5 6
0 0 1
−2
5
−3
−1
A = −8 17 −10
5 −10 6
Section 8.9
2